Question

Prove that:
$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\text{cosec}\theta$

Answer 1

We have,

LHS = $\frac{sin\theta }{1+cos\theta }+\frac{1+cos\theta }{sin\theta }$

$\Rightarrow$ LHS = $\frac{si{n}^2\theta +(1+cos\theta {)}^2}{sin\theta (1+cos\theta )}$

$\Rightarrow$ LHS = $\frac{si{n}^2\theta +1+2cos\theta +co{s}^2\theta }{sin\theta (1+cos\theta )}$

$\Rightarrow$ LHS = $\frac{(si{n}^2\theta +co{s}^2\theta )+1+2cos\theta }{sin\theta (1+cos\theta )}$ $[\because si{n}^2\theta +co{s}^2\theta =1]$

$\Rightarrow$ LHS = $\frac{2+2cos\theta }{sin\theta (1+cos\theta )}=\frac{2(1+cos\theta )}{sin\theta (1+cos\theta )}=\frac{2}{sin\theta }=2cosec\theta =RHS$